A division of a company has over $200$ employees, $40\%$ of which are male. The company is going to randomly select $4$ of these employees to attend a conference. Let $X=$ the number of male employees chosen. What is the probability that exactly $3$ of the $4$ employees chosen are male? You may round your answer to the nearest hundredth. $P(X=3)=$
Answer: Without a fancy calculator For $3$ of the $4$ employees chosen to be male, we needs to get $3$ males chosen and $1$ person who isn't male. We know $P({\text{male}})={40\%}$ and $P({\text{not}})={60\%}$. We can assume independence since we are sampling less than $10\%$ of the population. So let's multiply probabilities to find the probability of getting $3$ males followed by $1$ person who isn't male: $P({\text{MMM}}{\text{N}})=({0.4})^3({0.6})=0.0384$ This isn't our final answer, because there are other ways for $3$ of the $4$ employees chosen to be male (for example, NMMM). How many different ways are there? We can use the combination formula to find how many ways there are to get $3$ males in a sample of $4$ employees: $\begin{aligned} _n\text{C}_k&=\dfrac{n!}{(n-k)!\cdot k!} \\\\ _4\text{C}_3&=\dfrac{4!}{(4-3)!\cdot3!} \\\\ &=\dfrac{4 \cdot \cancel{3 \cdot 2 \cdot 1}}{(1) \cdot \cancel{3 \cdot 2 \cdot 1}} \\\\ &=4 \end{aligned}$ There are $4$ ways to get $3$ males in a sample of $4$ employees. Do they all have the same probability? Each of the $4$ ways has the same probability that we already found: $\begin{aligned} P({\text{MMM}}{\text{N}})&=({0.4})^3({0.6})=0.0384 \\\\ P({\text{MM}}{\text{N}}{\text{M}})&=({0.4})^3({0.6})=0.0384 \\\\ P({\text{M}}{\text{N}}{\text{MM}})&=({0.4})^3({0.6})=0.0384 \\\\ P({\text{N}}{\text{MMM}})&=({0.4})^3({0.6})=0.0384 \end{aligned}$ So we can multiply this probability by $4$ since that is how many ways there are to get $3$ males in a sample of $4$ employees: $\begin{aligned} P(X=3)&=4(0.4)^3(0.6) \\\\ &=4(0.0384) \\\\ &=0.1536 \end{aligned}$ Answer $P(X=3)=0.1536\approx0.15$